Calculate the pH at the equivalence volume for the titration of 25.0 mL of 0.350 M...
Homework Answers
pKa = - log Ka = - log (1.8 * 10^-5) = 4.745
volume of NaOH solution needed at equivalence point = 25.0 mL * 0.350 M / 0.350 M = 25.0 ml
and
total volume of solution = (25.0 + 25.0) = 50.0 ml
at equivalence point,
CH3COOH + NaOH ..........> CH3COONa + H2O
[salt] = 25.0 ml * 0.350 M / 50.0 ml = 0.175 M
so
salt hydrolysis occurs.
pH = 1/2 * [pKw + pKa + log C]
or
pH = 1/2 * [14 + 4.745 + log (0.175)]
or
pH = 8.994
or
pH = 9.0 (answer)
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